If R be a relation defined on the set of Natural numbers as " x R y (x-y) is divisible by 5 " x, y N then the relation R is

Option a: Equivalence. Solution: Let R be defined on the set of natural numbers as: x R y (x - y) is divisible by 5 Let's check the three properties of an equivalence relation: 1. Reflexive: For any x : x - x = 0 Since 0 is divisible by 5 ( 0 = 5 × 0 ), x R x holds. The relation is reflexive. 2. Symmetric: If x R y , then (x - y) is divisible by 5, i.e., x - y = 5k for some integer k . y - x = -(x - y) = -5k = 5(-k) Since -k is an integer, (y - x) is also divisible by 5, meaning y R x . The relation is symmetric. 3. Transitive: If x R y and y R z , then x - y = 5k1 and y - z = 5k2 for some integers k1 and k2 . Adding these: (x - y) + (y - z) = 5k1 + 5k2 x - z = 5(k1 + k2) Since k1 + k2 is an integer, (x - z) is divisible by 5, meaning x R z . The relation is transitive. Since R is reflexive, symmetric, and transitive, it is an Equivalence relation. Hence, Option A is the correct answer.

If $R$ be a relation defined on the set of Natural numbers as "$x$ $R$ $y \Leftrightarrow (x-y)$ is divisible by $5$" $\forall x, y \in N$ then the relation $R$ is

The correct answer is Option a: Equivalence. Let $R$ be defined on the set of natural numbers $\mathbb{N}$ as: $$x R y \iff (x - y) \text{ is divisible by } 5$$ Let's check the three properties of an equivalence relation: 1. **Reflexive**: For any $x \in \mathbb{N}$: $$x - x = 0$$ Since $0$ is divisible by 5 ($0 = 5 \times 0$), $x R x$ holds. The relation is reflexive. 2. **Symmetric**: If $x R y$, then $(x - y)$ is divisible by 5, i.e., $x - y = 5k$ for some integer $k$. $$y - x = -(x - y) = -5k = 5(-k)$$ Since $-k$ is an integer, $(y - x)$ is also divisible by 5, meaning $y R x$. The relation is symmetric. 3. **Transitive**: If $x R y$ and $y R z$, then $x - y = 5k_1$ and $y - z = 5k_2$ for some integers $k_1$ and $k_2$. Adding these: $$(x - y) + (y - z) = 5k_1 + 5k_2$$ $$x - z = 5(k_1 + k_2)$$ Since $k_1 + k_2$ is an integer, $(x - z)$ is divisible by 5, meaning $x R z$. The relation is transitive. Since $R$ is reflexive, symmetric, and transitive, it is an **Equivalence** relation. Hence, **Option A** is the correct answer.

Sets, Relations and FunctionsMCQPYQ Jun 23Question 1947 of 217

All Questions

If $\displaystyle R$ be a relation defined on the set of Natural numbers as " $\displaystyle x$ $\displaystyle R$ $\displaystyle y \Leftrightarrow (x-y)$ is divisible by $\displaystyle 5$ " $\displaystyle \forall x, y \in N$ then the relation $\displaystyle R$ is

Options

AEquivalence

BAnti-symmetric

CSymmetric but not transitive

DSymmetric but not reflexive

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Correct Answer

✅ Option a — Equivalence

All Options:

AEquivalence
BAnti-symmetric
CSymmetric but not transitive
DSymmetric but not reflexive

Detailed Solution & Explanation

Let

\displaystyle R

be defined on the set of natural numbers

\displaystyle \mathbb{N}

as:

x R y \iff (x - y) \text{ is divisible by } 5

Let's check the three properties of an equivalence relation:
1. **Reflexive**: For any

\displaystyle x \in \mathbb{N}

x - x = 0

Since

\displaystyle 0

is divisible by 5 (

\displaystyle 0 = 5 \times 0

\displaystyle x R x

holds. The relation is reflexive.
2. **Symmetric**: If

\displaystyle x R y

, then

\displaystyle (x - y)

is divisible by 5, i.e.,

\displaystyle x - y = 5k

for some integer

\displaystyle k

y - x = -(x - y) = -5k = 5(-k)

Since

\displaystyle -k

is an integer,

\displaystyle (y - x)

is also divisible by 5, meaning

\displaystyle y R x

. The relation is symmetric.
3. **Transitive**: If

\displaystyle x R y

and

\displaystyle y R z

, then

\displaystyle x - y = 5k_1

and

\displaystyle y - z = 5k_2

for some integers

\displaystyle k_1

and

\displaystyle k_2

. Adding these:

(x - y) + (y - z) = 5k_1 + 5k_2

x - z = 5(k_1 + k_2)

Since

\displaystyle k_1 + k_2

is an integer,

\displaystyle (x - z)

is divisible by 5, meaning

\displaystyle x R z

. The relation is transitive.
Since

\displaystyle R

is reflexive, symmetric, and transitive, it is an **Equivalence** relation.
Hence, **Option A** is the correct answer.

About This Chapter: Sets, Relations and Functions

Paper

Paper 3: Quantitative Aptitude

Weightage

3-5 Marks

Key Topics

Sets, Relations, Functions

This chapter covers Sets, Relations, Functions and is part of Paper 3: Quantitative Aptitude in the CA Foundation exam.

View Official ICAI Syllabus

Exam Strategy Tip

This topic carries 3-5 Marks weightage. Focus on understanding core concepts rather than memorizing.

Key Concepts to Understand

Set

A well-defined collection of distinct objects (called elements or members). Sets are denoted by capital letters and elements are listed within curly braces. Key operations include Union (∪), Intersection (∩), and Complement.

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If R\displaystyle RR be a relation defined on the set of Natural numbers as "x\displaystyle xx R\displaystyle RR y⇔(x−y)\displaystyle y \Leftrightarrow (x-y)y⇔(x−y) is divisible by 5\displaystyle 55" ∀x,y∈N\displaystyle \forall x, y \in N∀x,y∈N then the relation R\displaystyle RR is

Options

Correct Answer

All Options:

Detailed Solution & Explanation

About This Chapter: Sets, Relations and Functions

Key Concepts to Understand

Set

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Ready to Master Sets, Relations and Functions?

If $\displaystyle R$ be a relation defined on the set of Natural numbers as " $\displaystyle x$ $\displaystyle R$ $\displaystyle y \Leftrightarrow (x-y)$ is divisible by $\displaystyle 5$ " $\displaystyle \forall x, y \in N$ then the relation $\displaystyle R$ is